It would take greater than 31.25 hours for Tim's price to be better. So in terms of whole hours it would take at least 32 hours of work for Tim's price to be better.
It has 3 zeros, because its a polynomial of degree 3.
Answer: The inverse of the linear function f(x)=2x+1 is f^(-1) (x) = (1/2)x-1/2
Solution
f(x)=2x+1
y=f(x)
y=2x+1
Isolating x: Subtracting 1 both sides of the equation:
y-1=2x+1-1
y-1=2x
Multiplying both sides of the equation by 1/2:
(1/2)(y-1)=(1/2)2x
(1/2)y-1/2=x
x=(1/2)y-1/2
Changing "x" by "f^(-1) (x)" and "y" by "x":
f^(-1) (x) = (1/2)x-1/2
THIS IS THE COMPLETE QUESTION BELOW
The demand equation for a product is p=90000/400+3x where p is the price (in dollars) and x is the number of units (in thousands). Find the average price p on the interval 40 ≤ x ≤ 50.
Answer
$168.27
Step by step Explanation
Given p=90000/400+3x
With the limits of 40 to 50
Then we need the integral in the form below to find the average price
1/(g-d)∫ⁿₐf(x)dx
Where n= 40 and a= 50, then if we substitute p and the limits then we integrate
1/(50-40)∫⁵⁰₄₀(90000/400+3x)
1/10∫⁵⁰₄₀(90000/400+3x)
If we perform some factorization we have
90000/(10)(3)∫3dx/(400+3x)
3000[ln400+3x]₄₀⁵⁰
Then let substitute the upper and lower limits we have
3000[ln400+3(50)]-ln[400+3(40]
30000[ln550-ln520]
3000[6.3099×6.254]
3000[0.056]
=168.27
the average price p on the interval 40 ≤ x ≤ 50 is
=$168.27
Answer:
373.8mmHg
Step-by-step explanation:
a =height (in km) above sea level,
the pressure P(a) (in mmHg) is approximated given as
P(a) = 760e–0.13a .
To determine the atmospheric pressure at 5.458 km, then we will input into the equation
P(5.458km) = 760e–0.13a .
= 760e^(-0.13×5.458)
=760e^-(0.70954)
= 760×0.4919
=373.8mmHg
Therefore, the atmospheric pressure at 5.458 km is 373.8mmHg
